Chapter 15 - Section 15.3 - Maxima and Minima - Exercises - Page 1117: 17

$${\text{Relative minimum at the point }}\left( {4, - 3, - 10} \right)$$

$$\eqalign{ & f\left( {x,y} \right) = {x^2} + 2xy + 2{y^2} - 2x + 4y \cr & {\text{Calculate the first partial derivatives of }}f\left( {x,y} \right) \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{x^2} + 2xy + 2{y^2} - 2x + 4y} \right] \cr & {f_x}\left( {x,y} \right) = 2x + 2y - 2 \cr & and \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{x^2} + 2xy + 2{y^2} - 2x + 4y} \right] \cr & {f_y}\left( {x,y} \right) = 2x + 4y + 4 \cr & {\text{Setting both first partial derivatives equal to zero, we have}} \cr & {f_x}\left( {x,y} \right) = 0,{\text{ }}{f_y}\left( {x,y} \right) = 0 \cr & 2x + 2y - 2 = 0{\text{ }}\left( {\bf{1}} \right),{\text{ }}2x + 4y + 4 = 0{\text{ }}\left( {\bf{2}} \right) \cr & {\text{Solving the equations simultaneously, we obtain}} \cr & x = 4,{\text{ }}y = - 3 \cr & {\text{The critical point is }}\left( {4, - 3} \right) \cr & {\text{Find the second partial derivatives of }}f\left( {x,y} \right){\text{ and }}{f_{xy}}\left( {x,y} \right) \cr & {f_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {2x + 2y - 2} \right] = 2 \cr & {f_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {2x + 4y + 4} \right] = 4 \cr & {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {2x + 2y - 2} \right] = 2 \cr & {\text{By the second partials test}} \cr & d\left( {x,y} \right) = {f_{xx}}\left( {a,b} \right){f_{yy}}\left( {a,b} \right) - {\left[ {{f_{xy}}\left( {x,y} \right)} \right]^2} \cr & {\text{Evaluate }}d\left( {x,y} \right){\text{ at the critical point }}\left( {4, - 3} \right) \cr & d\left( {4, - 3} \right) = {f_{xx}}\left( {4, - 3} \right){f_{yy}}\left( {4, - 3} \right) - {\left[ {{f_{xy}}\left( {4, - 3} \right)} \right]^2} \cr & d\left( {4, - 3} \right) = \left( 2 \right)\left( 4 \right) - {\left[ 2 \right]^2} \cr & d\left( {4, - 3} \right) = 4 \cr & d > 0,{\text{ and }}{f_{xx}}\left( {4, - 3} \right) = 2 > 0 \cr & {\text{then}} \cr & f\left( {x,y} \right){\text{ has a relative minimum at }}\left( {4, - 3,f\left( {4, - 3} \right)} \right) \cr & f\left( {x,y} \right) = {x^2} + 2xy + 2{y^2} - 2x + 4y \cr & f\left( {4, - 3} \right) = {\left( 4 \right)^2} + 2\left( 4 \right)\left( { - 3} \right) + 2{\left( { - 3} \right)^2} - 2\left( 4 \right) + 4\left( { - 3} \right) \cr & f\left( {4, - 3} \right) = - 10 \cr & {\text{Relative minimum at the point }}\left( {4, - 3, - 10} \right) \cr} $$